Question

Crude oil biodegradation. Refer to the Journal of Petroleum Geology (April 2010) study of the environmental factors associated with biodegradation in crude oil reservoirs, Exercise 6.38 (p. 350). Recall that 16 water specimens were randomly selected from various locations in a reservoir on the floor of a mine and that the amount of dioxide (milligrams/liter)—a measure of biodegradation—as well as presence of oil were determined for each specimen. These data are reproduced in the accompanying table.

a. Conduct a test to determine if the true mean amount of dioxide present in water specimens that contained oil was less than 3 milligrams/liter. Use\(\alpha = .10\)

Short answer

1. The test statistic is -14.14
   We reject the null hypothesis

Step-by-step solution

Specifying the hypothesis

The null and alternative hypothesis are given by

\(\begin{aligned}{H_0}:\mu = 3\\{H_a}:\mu < 3\end{aligned}\)

calculating the mean and standard deviation

a) The mean and standard deviation is calculated as

\(\begin{aligned}\bar x &= \frac{{0.5 + 1.3 + 0.4 + 0.2 + 0.5 + 0.2}}{6}\\ &= \frac{{3.1}}{6}\\ &= 0.51\end{aligned}\)

\(\begin{aligned}sd &= \sqrt {\frac{{{{\left( { - 0.01} \right)}^2} + {{\left( {0.79} \right)}^2} + {{\left( { - 0.11} \right)}^2} + {{\left( { - 0.31} \right)}^2} + {{\left( { - 0.01} \right)}^2} + {{\left( { - 0.31} \right)}^2}}}{5}} \\ &= \sqrt {\frac{{0.0001 + .6241 + 0.0121 + 0.0961 + 0.0001 + 0.0961}}{5}} \\ &= \sqrt {\frac{{0.9375}}{5}} \\ &= 0.433\end{aligned}\)

Therefore, the mean and standard deviation are 0.51 and 0.433.

Test statistic

The test statistic is calculated as

\(\begin{aligned}t &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{0.51 - 3}}{{\frac{{0.433}}{{\sqrt 6 }}}}\\ &= \frac{{ - 2.49}}{{0.176}}\\ &= - 14.14\end{aligned}\)

Therefore, the test statistic is -14.14.

Degrees of freedom are

\(\begin{aligned}df &= n - 1\\ &= 6 - 1\\ &= 5\end{aligned}\)

For \(\alpha = .10\,and\,df = 5\)

The tabulated value is -1.47.

The calculated value is less than the tabulated value.

Therefore, we reject the null hypothesis.